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# Help

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Sep 30, 2019
edited by SoulSlayer615  Oct 23, 2019

#1
+2547
+2

a) Hint: Plus it in

$$\sqrt{5^2+12^2}$$, then solve. You should get an integer as an answer

b) To do this problem, you can square both sides.

$$x^2+2xy+y^2=x^2+y^2$$

Then cancel out x^2 and y ^2

$$2xy = 0$$

Then solve

$$x*y=0$$.

So when x and y multiplied together equaling 0 is when it can satisfy the original equation.

Examples: (0, 0)  (10, 0)    (0, 5)

Values of x and y that DO NOT satisfy the original equation is when they DO NOT multiply to 0

Examples (1, 1) (2, 3) (6, 9)

Sep 30, 2019
#2
+999
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I just graphed it, you can find it out!

Oct 1, 2019
#3
+2547
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Yes, when graphed, it will always end up on the axis lines!

CalculatorUser  Oct 1, 2019
#4
+106539
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x + y  =  sqrt (x^2 + y^2)      square both sides

x^2 + 2xy + y^2  =  x^2 + y^2

2xy  = 0

xy  =  0

Then.....this will be true as long as either x,y  = 0   or both = 0

Similarly

x + y  will not equal sqrt (x^2 + y^2)    whenever x and y are both non-zero

Oct 1, 2019